Daniil wrote an excellent post, but just to add to that a little bit:
As Daniil pointed out, you can't capture any truth-functions that non-trivially depend on more than $1$ variable, such as $P \land Q$, with only a $\neg$. So, let's restrict ourselves to functions defined over one variable, $P$, and see if maybe we can capture all those using a $\neg$?
Unfortunately, the answer is still no. Again, as Daniil already pointed out, we can't capture any tautology or contradiction. That is, we can't capture the truth-function that always returns true (i.e. the function $f$ such that $f(T)=f(F)=T$), nor can we capture the truth-function that always returns false (i.e. the function $f'$ such that $f'(T)=f'(F)=F$)
So in this post I just wanted to show you how you can prove that result using induction. In particular, let's prove the following:
Claim
For any expression $\phi$ built up from $P$ and $\neg$ alone, it will be true that if $v$ is the valuation that sets $P$ to true (i.e. $v(P)=T$), and $v'$ is the valuation that sets $P$ to false (i.e. $v'(P)=F$), then either $v(\phi)=T$ and $v'(\phi)=F$, or $v'(\phi)=T$ and $v(\phi)=F$ (in other words, $v(\phi)$ and $v'(\phi)$ will always opposite values, meaning that $\phi$ can not be a tautology or contradiction, for that would require that $\phi$ has the same value for any valuation)
Proof
We'll prove the claim by structural induction on the formation of $\phi$:
*Base: *
$\phi=P$. Then $v(\phi)=v(P)=T$, while $v'(\phi)=v'(P)=F$. Check!
Step:
If $\phi$ is not an atomic proposition, then there is only one possibility: $\phi$ is the negation of some other statement $\psi$, i.e. $\phi = \neg \psi$.
Now, by inductive hypothesis we can assume that $v(\psi)=T$ and $v'(\psi)=F$, or $v'(\psi)=T$ and $v(\psi)=F$
Well, if $v(\psi)=T$ and $v'(\psi)=F$, then $v(\phi)=v(\neg \psi)=F$ and $v'(\phi)=v'(\neg \psi) =T$. On the other hand, if $v(\psi)=F$ and $v'(\psi)=T$, then $v(\phi)=v(\neg \psi)=T$ and $v'(\phi)=v'(\neg \psi) =F$. So, we can conclude that $v(\phi)=T$ and $v'(\phi)=F$, or $v'(\phi)=T$ and $v(\phi)=F$, as desired.
No, that's not a valid argument. The problem is when you say
But this says that one truth assignment which satisfies $\Sigma$ make none of $p_i$, $i\in\mathbb{N}$ true
That's not what it says. You have for each $n$ a valuation $\nu_n$ which makes $\Sigma$ true but no $p_i$ true for $i\le n$, but you haven't cooked up a single valuation $\nu$ which makes $\Sigma$ true but no $p_i$ true for any $i$ at all.
For example, maybe $\nu_1\models p_2$, $\nu_2\models p_3$, $\nu_3\models p_4$, ...
Now we want to say that somehow we can "smoosh together" the $\nu_i$s to get a $\nu$ making $\Sigma$ true and making each $p_i$ false. But how?
(HINT: use the compactness theorem ...)
(We can sometimes think of this as "taking the limit" of the $\nu_n$s, but that takes a bit of work to make non-problematic. One approach to doing so is to introduce ultrafilters - and this is basically avoiding using the compactness theorem by employing its proof.)
Best Answer
Your question is slightly ambiguous. In my opinion, there are two possible interpretations of your question:
Let $V$ be the set of all propositional variables (they are countably infinitely many). Then, $V$ is consistent and complete but not maximally consistent.
Proof.
Consistency: Clearly, there is a model $v$ of $V$, the one such that $v(P) = \top$ for all $P \in V$. By soundness theorem, $V$ is consistent i.e. there is no formula $\varphi$ such that $V \vdash \varphi$ and $V \vdash \lnot \varphi$.
No maximal consistency: As you correctly said, for any $P, Q \in V$, one has $V \vdash P \land Q$ since $V \vdash P$ and $V \vdash Q$, but the formula $P \land Q \notin V$.
Completeness: Let $\varphi$ be a formula. There is only one truth assignment $v$ that is a model of $V$: it is the truth assignment $v$ such that $v(P) = \top$ for all $P \in V$. As any truth assignment, either $v \vDash \varphi$ or $v \vDash \lnot \varphi$. Hence, either $V \vDash \varphi$ or $V \vDash \lnot \varphi$, i.e. either all the models of $V$ are a model of $\varphi$ or all the models of $V$ are a model of $\lnot \varphi$. By completeness theorem, $V \vdash \varphi$ or $V \vdash \lnot \varphi$. $\quad\square$
Let $V$ be a proper subset of the set of all propositional variables (in particular, this is the case if $V$ is a finite set of propositional variables). Then, $V$ is consistent but neither complete nor maximal consistent.
Proof.
Consistency: As above.
No maximal consistency: given a propositional variable $P \notin V$ (it exists by hypothesis), $V \cup \{P\}$ is consistent (see consistency in Point 1), but $V \subsetneq V \cup \{P\}$.
No completeness: As correctly suggested by Asaf Kasaglia's comment, given a propositional variable $P \notin V$ (it exists by hypothesis), from the consistency of $V \cup \{P\}$ it follows that $V \not\vdash \lnot P$. But $V \cup \{\lnot P\}$ is consistent as well (take the truth assignment $v$ such that $v(Q) = \top$ for all $Q \in V$, and $v(P) = \bot$, then use soundness theorem to conclude), hence $V \not \vdash P$. $\qquad \square$